Question

present ages of x and y are in the ratio 7:12.if two years ago the ratio of their age was 3:8,their present ages are​

Answer 1

Answer:

let

the present ages 7x,12x

2 years ago 3y,8y

7x-3y=2

12x-8y=2

solve the equation,x=1/2,y=1/2

the present age of x and y are 1.5,6

Answer 2

⇝Given :

• ➵ Present age of x and y are in the ratio 7:12 .2 years ago the ratio of their ages was 3:8 .

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⇝To Find :

• ➵ Present age of x and y = ?

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⇝Solution :

★ Present age of x and y are in the ratio 7:12 .Hence,

${\qquad{\large{\gray{\bigstar}} \: \:{\underline{\underline{\orange{\sf{ \dfrac{x}{y} = \dfrac{7x}{12x}}}}}}}}$

★ Two years ago their ages were in the ratio 3:8 .Hence,

${\qquad{\large{\gray{\bigstar}} \: \:{\underline{\underline{\orange{\sf{ \dfrac{x}{y} = \dfrac{7x - 2}{12x - 2} = \dfrac{3}{8}}}}}}}}$

$\qquad{━━━━━━━━━━━━━━━━━━━━}$

~ Now Cross Multiplication :

${:\longmapsto{\qquad{\sf{ \dfrac{7x - 2}{12x - 2} = \dfrac{3}{8}}}}}$

${:\longmapsto{\qquad{\sf{ 8(7x - 2) = 3(12x - 2)}}}}$

${:\longmapsto{\qquad{\sf{ 56x - 16 = 36x - 6}}}}$

${:\longmapsto{\qquad{\sf{ 56x - 36x = 16 - 6}}}}$

${:\longmapsto{\qquad{\sf{ 20x = 10}}}}$

${:\longmapsto{\qquad{\sf{ x = \dfrac{10}{20}}}}}$

${:\longmapsto{\qquad{\sf{ x = \cancel\dfrac{10}{20}}}}}$

${\qquad{\sf{ Value \: of \: x \: = {\color{maroon}{\sf{ 0.5 }}}}}}$

~ Calculating the Ages :

$\leadsto{\qquad{\pink{\sf{ Age \: of \: x = 7x = 7 \times 0.5 = {\blue{\sf{3.5 \: years }}}}}}}$

$\leadsto{\qquad{\pink{\sf{ Age \: of \: y = 12x = 12 \times 0.5 = {\blue{\sf{6 \: years }}}}}}}$

$\qquad{━━━━━━━━━━━━━━━━━━━━}$

Therefore :

❝ Age of x is ${\pink{\underline{\color{purple}{\sf{3.5 \: years}}}}}$ .❞

❝ Age of y is ${\pink{\underline{\color{purple}{\sf{6 \: years}}}}}$ .❞

${\orange{\underline{\red{▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬}}}}$